Question

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100\(M \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{g} \mathrm{com}-\) pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)

Short answer

The formula of compound A is approximately Cr(NH₃)₂₅I₅. The structure of the complex ion present in compound A is likely [Cr(NH₃)₆]³⁺ or a similar complex ion containing chromium with six-coordinate ammonia ligands.

Step-by-step solution

Determine the moles of CrO₃ formed

We are given that 0.105 g of compound A was strongly heated in excess O₂, and 0.0203 g CrO₃ was formed. To find the moles of CrO₃, we use the molar mass of CrO₃ (151.99 g/mol): Moles of CrO₃ = mass / molar mass Moles of CrO₃ = 0.0203 g / 151.99 g/mol = 1.33 x 10⁻⁴ moles

Calculate the moles of Cr present in compound A

Since only one Chromium atom is present in CrO₃, there is an equal number of moles of Chromium in compound A: Moles of Cr = 1.33 x 10⁻⁴ moles

Calculate the moles of NH₃ present in compound A

It took 32.93 mL of 0.100 M HCl to titrate completely the NH₃ present in 0.341 g compound A. We can calculate the moles of NH₃ in compound A by using the volume and concentration of HCl: Moles of NH₃ = 32.93 mL * 0.100 mol/L = 3.293 x 10⁻³ moles

Calculate the moles of I present in compound A

Compound A was found to contain 73.53 % iodine by mass. We can use this percentage to calculate the mass of iodine in 0.105 g of compound A: Mass of I = 0.105 g * 0.7353 = 0.0772 g To find the moles of I, we must use the molar mass of I (126.90 g/mol): Moles of I = 0.0772 g / 126.90 g/mol = 6.09 x 10⁻⁴ moles

Determine the molecular formula of compound A

To determine the molecular formula of compound A, we need to find the ratio of moles of Cr, NH₃, and I in the compound. To simplify, we will divide each mole value by the smallest value (1.33 x 10⁻⁴ moles Cr): - Cr: (1.33 x 10⁻⁴) / (1.33 x 10⁻⁴) = 1 - NH₃: (3.293 x 10⁻³) / (1.33 x 10⁻⁴) = 24.74 ≈ 25 - I: (6.09 x 10⁻⁴) / (1.33 x 10⁻⁴) = 4.58 ≈ 5 Therefore, the molecular formula of compound A is approximately Cr(NH₃)₂₅I₅.

Determine the structure of the complex ion present

We are given a hint that Cr³⁺ is expected to be six-coordinate, with NH₃ and possibly I⁻ as ligands. The I⁻ ions will be the counterions if needed. Since there are a large number of ammonia ligands (25 of them) in the molecular formula, it seems likely that the complex ion present in compound A contains Cr³⁺ surrounded by six ammonia ligands, forming [Cr(NH₃)₆]³⁺. The I⁻ ions will provide charge balance, and the resulting formula would be [Cr(NH₃)₆]I₃, but this formula doesn't conform fully to the stoichiometry found in step 5. The specific arrangement of ammonia ligands and iodide ions in compound A may not be easily determined, but we can infer that the complex ion in compound A is [Cr(NH₃)₆]³⁺ or a similar complex ion containing chromium with six-coordinate ammonia ligands.

Key concepts

Chromium Coordination Chemistry

Chromium (Cr) is a fascinating transition metal known for its ability to form complex ions with various ligands. In the context of chromium coordination chemistry, Cr often adopts a coordination number of six, meaning it can bind to six other atoms or ions. This is due to its electronic configuration, which provides enough space to accommodate multiple ligands.

The electronic configuration of chromium in its common oxidation state, Cr³⁺, is \[\text{[Ar]} 3d^3\]. This allows it to engage in strong interactions with ligands, particularly those that can donate lone pairs of electrons, such as ammonia (NH₃) and iodide (I⁻).

In compound A from our exercise, Cr³⁺ likely forms a complex ion with ammonia as its primary ligand. The Cr-N bond is largely ionic with covalent character provided by the lone pair donation from NH₃ to the empty d orbitals of chromium. Understanding these interactions is crucial in predicting the structure and properties of chromium complexes.

Ligand Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. But in coordination chemistry, ligand stoichiometry involves understanding the ratios of ligands to central metal atoms within a complex.

In the exercise, compound A shows a surprisingly large number of ammonia ligands, much higher than one would initially expect. Despite step-by-step analyses determining a formula like Cr(NH₃)₂₅I₅, coordination chemistry typically adheres to more fixed ligand limits in known stable complexes.

Commonly, chromium complexes conform to a 1:6 metal-to-ligand ratio, as hinted by the six-coordinate model for Cr³⁺. The analysis suggests that while many ammonia molecules interact with the structure, only six coordinate directly with Cr in the core complex ion [Cr(NH₃)₆]³⁺.

The presence of excess ammonia could be part of an external complex structure or intermolecular interactions, serving different roles like stabilization, which was not fully captured in the molecular formula originally considered.

Complex Ion Formation

Complex ion formation is a key phenomenon in transition metal chemistry, especially for elements like chromium. A complex ion consists of a central metal ion bonded to surrounding molecules or ions called ligands.

In forming complex ions, chromium shows a preference for maintaining a coordination number of six, as in [Cr(NH₃)₆]³⁺. This reflects the metal's typical coordination behavior influenced by crystal field theory, which helps predict complex formation based on ligand field strengths.

The ligand ammonia in the exercise behaves as a neutral donor, stabilizing the metal center effectively through donation of electron pairs. On the other hand, iodide ions, while potentially acting as ligands, in this context are likely counterions providing charge neutrality outside the immediate coordination sphere of the chromium.

This understanding of complex ion formation allows chemists to predict properties and synthesize new compounds with desired chemical and physical characteristics. Through practice, students can better appreciate the intricate dance of electrons and the synergy between metal ions and ligands that bring complex ions to life.